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suter [353]
3 years ago
7

The probability that a single radar station will detect an enemy plane is 0.65. (a) How many such stations are required to be 98

% certain that an enemy plane flying over will be detected by at least one station?
Mathematics
1 answer:
Inga [223]3 years ago
4 0

Answer:

4 stations

Step-by-step explanation:

If we need to be at least 98% certain that an enemy plane flying over will be detected by at least one station, we must ensure that there is at most a 2% chance that no radar stations detect the plane.

The probability that a single radar station does not detect the plane is 0.35.

For n radar stations:

P=0.02=0.35^n\\n=\frac{log(0.02)}{log(0.35)}\\n=3.73

Rounding up to the next whole station, at least 4 stations are needed.

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Answer:

95.64% probability that pledges are received within 40 days

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 28, \sigma = 7

What is the probability that pledges are received within 40 days

This is the pvalue of Z when X = 40. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{40 - 28}{7}

Z = 1.71

Z = 1.71 has a pvalue of 0.9564

95.64% probability that pledges are received within 40 days

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For this case we have that by definition, the density is given by:

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